Q.Integrate the following function: $e^x \left(\frac{1+\sin x}{1+\cos x}\right)$

Showing the 12 most recent of 51 on this concept.

1. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{x^7+x^{11}}{(1+2x^4-5x^8)^2}\,dx = \dfrac{x^n}{8(1+2x^4-5x^8)} + c$, then $n =$ (A) $10$ (B) $8$ (C) $16$ (D) $12$

   ›Reveal solutionSolution

   Differentiate the proposed antiderivative form with unknown power $n$, expand, and match coefficients against the given integrand $(x^7+x^{11})/D^2$ — this pins down $n=8$.

   Concept and Intuition

   When an integral is given already in "answer form" with an unknown parameter, the fastest and most reliable check is reverse engineering: differentiate the claimed antiderivative and see what power $n$ makes it reproduce the original integrand exactly.

   Step-by-Step Solution

   1. Let $D(x) = 1+2x^4-5x^8$, so $D'(x) = 8x^3-40x^7$.
   2. We're told $\displaystyle\int\dfrac{x^7+x^{11}}{D(x)^2}\,dx = \dfrac{x^n}{8D(x)}+c$. Differentiate the RHS using the quotient rule: $\dfrac{d}{dx}\left[\dfrac{x^n}{8D}\right] = \dfrac{1}{8}\cdot\dfrac{nx^{n-1}D - x^n D'}{D^2}$.
   3. This must equal $\dfrac{x^7+x^{11}}{D^2}$, so: $nx^{n-1}D - x^nD' = 8(x^7+x^{11})$.
   4. Try $n=8$: $8x^7\cdot D - x^8\cdot D' = 8x^7(1+2x^4-5x^8) - x^8(8x^3-40x^7)$ $= 8x^7 + 16x^{11} - 40x^{15} - 8x^{11} + 40x^{15}$ $= 8x^7 + (16-8)x^{11} + (-40+40)x^{15} = 8x^7 + 8x^{11}$. …

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2. AP EAPCET 2026Set eng-2026-05-12-FN1 markMCQ
   Q.If $g(x)$ and $f(x)$ are twice differentiable functions, then $\displaystyle\int (f''(x)g(x) - g''(x)f(x))\,dx =$ (A) $g^2(x)\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)$ (B) $f^2(x)\dfrac{d}{dx}\left(\dfrac{f'(x)}{g'(x)}\right)$ (C) $\dfrac{d}{dx}\left(\dfrac{f''(x)}{g''(x)}\right)g^2(x)$ (D) $\dfrac{d}{dx}\left(\dfrac{f'(x)}{g(x)}\right)g^2(x)$

   ›Reveal solutionSolution

   The integrand $f''g-g''f$ is exactly the derivative of $f'g-fg'$, which in turn is $g^2$ times the derivative of the quotient $f/g$ — a classic "spot the exact derivative" integration trick.

   Concept and Intuition

   Rather than integrating term by term, recognize that many such expressions are secretly the derivative of a simpler combination. Here, differentiating $f'g-fg'$ reproduces the given integrand exactly, so no actual "integration" work is needed — just pattern recognition, followed by rewriting the result using the quotient rule identity.

   Step-by-Step Solution

   1. Consider the candidate antiderivative $h(x) = f'(x)g(x) - f(x)g'(x)$.
   2. Differentiate: $h'(x) = \left[f''(x)g(x)+f'(x)g'(x)\right] - \left[f'(x)g'(x)+f(x)g''(x)\right]$ $= f''(x)g(x) - g''(x)f(x)$ (the $f'g'$ terms cancel).
   3. This is exactly the given integrand! So $\displaystyle\int\left(f''(x)g(x)-g''(x)f(x)\right)dx = f'(x)g(x)-f(x)g'(x) + c$.
   4. Now recall the quotient rule: $\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right) = \dfrac{f'(x)g(x)-f(x)g'(x)}{g^2(x)}$.
   5. So $f'(x)g(x)-f(x)g'(x) = g^2(x)\cdot\dfrac{d}{dx}\left(\dfrac{f(x)}{g(x)}\right)$. …

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3. AP EAPCET 2026Set eng-2026-05-13-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{2\sin x + a\cos x}{b\sin x + 4\cos x}\,dx = \dfrac{2}{5}x - \dfrac{1}{5}\log(b\sin x + 4\cos x) + c$, then $a + b =$ (A) $2$ (B) $3$ (C) $4$ (D) $5$

   ›Reveal solutionSolution

   Split the numerator as $A(\text{denom})+B(\text{denom}')$ so the integral becomes

   $Ax+B\log|\text{denom}|$; matching this to the given $A=2/5$, $B=-1/5$ pins down

   $b=3$ and $a=1$, so $a+b=4$.

   Concept and Intuition

   Any integral of the form $\int\dfrac{p\sin x+q\cos x}{r\sin x+s\cos x}\,dx$ can always be

   written as $Ax + B\log|r\sin x+s\cos x| + c$, because

   $p\sin x+q\cos x$ can always be decomposed uniquely as $A(r\sin x+s\cos x) + B(r\cos x - s\sin x)$ — a multiple of the denominator plus a multiple of its derivative.

   Step-by-Step Solution

   1. Let denom $=b\sin x+4\cos x$; its derivative is $b\cos x-4\sin x$.
   2. Write $2\sin x+a\cos x = A(b\sin x+4\cos x)+B(b\cos x-4\sin x)$.
   3. Matching $\sin x$: $2=Ab-4B$. Matching $\cos x$: $a=4A+Bb$.
   4. The integral of $A+\dfrac{B\cdot\text{denom}'}{\text{denom}}$ is $Ax+B\log|\text{denom}|+c$; comparing to the given $\tfrac25x-\tfrac15\log(\cdot)+c$ gives $A=\tfrac25$, $B=-\tfrac15$. …

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4. AP EAPCET 2026Set eng-2026-05-14-AN1 markMCQ
   Q.If $\displaystyle\int \dfrac{2\cos x+3\sin x}{3\cos x+4\sin x}\,dx = Ax+B\log|3\cos x+4\sin x|+c$, then $A\cdot B=$ (A) $-\dfrac{18}{625}$ (B) $\dfrac{18}{625}$ (C) $-\dfrac{18}{25}$ (D) $\dfrac{18}{25}$

   ›Reveal solutionSolution

   This is the standard "express numerator as a·(denominator) + b·(derivative of denominator)" trick for $\int \dfrac{a\cos x+b\sin x}{c\cos x+d\sin x}dx$; solving the linear system gives $A\cdot B=-\dfrac{18}{625}$.

   Concept and Intuition

   Whenever the integrand is a ratio of two linear combinations of $\sin x,\cos x$, the denominator's own derivative (also a combination of $\sin x,\cos x$) together with the denominator itself spans the same 2-D space as the numerator. So we can always write numerator $= A\cdot(\text{denominator}) + B\cdot(\text{denominator})'$, turning the integral into $Ax + B\log|\text{denominator}|+c$ directly.

   Step-by-Step Solution

   1. Let $D=3\cos x+4\sin x$, so $D'=-3\sin x+4\cos x$.
   2. Write $2\cos x+3\sin x = A\,D + B\,D' = A(3\cos x+4\sin x)+B(-3\sin x+4\cos x)$.
   3. Match $\cos x$ coefficients: $3A+4B=2$.
   4. Match $\sin x$ coefficients: $4A-3B=3$.
   5. Solve: multiply the first by $3$ and the second by $4$: $9A+12B=6$, $16A-12B=12$. Adding, $25A=18\Rightarrow A=\dfrac{18}{25}$. …

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5. AP EAPCET 2026Set eng-2026-05-13-AN1 markMCQ
   Q.$\displaystyle\int \left(\dfrac{\log x - 1}{1+(\log x)^2}\right)^2 dx=$ (A) $\dfrac{x e^x}{1+x^2}+c$ (B) $\dfrac{x}{1+(\log x)^2}+c$ (C) $\dfrac{\log x}{(\log x)^2+1}+c$ (D) $\dfrac{x}{x^2+1}+c$

   ›Reveal solutionSolution

   The integrand is recognized as the derivative of $\dfrac{x}{1+(\log x)^2}$ via the quotient rule, so the antiderivative is that expression directly.

   Concept and Intuition

   Many integrals involving $\log x$ in a rational combination are disguised derivatives of a simple quotient $\dfrac{x}{P(\log x)}$. The strategy is to guess a plausible antiderivative form (informed by the structure of the options) and verify by differentiating.

   Step-by-Step Solution

   1. Guess $g(x) = \dfrac{x}{1+(\log x)^2}$ (matching option B's form).
   2. Differentiate using the quotient rule: $g'(x) = \dfrac{1\cdot(1+(\log x)^2) - x\cdot 2\log x \cdot \frac{1}{x}}{(1+(\log x)^2)^2}$.
   3. Simplify numerator: $(1+(\log x)^2) - 2\log x = 1 - 2\log x + (\log x)^2 = (\log x - 1)^2$.
   4. So $g'(x) = \dfrac{(\log x-1)^2}{(1+(\log x)^2)^2} = \left(\dfrac{\log x - 1}{1+(\log x)^2}\right)^2$, exactly the integrand. …

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6. AP EAPCET 2026Set eng-2026-05-15-AN1 markMCQ
   Q.$\int \frac{x-1}{(x+1)^3}e^x dx =$ (A) $\frac{e^x}{(x+1)^2} + c$ (B) $\frac{-e^x}{(x+1)^2} + c$ (C) $\frac{2e^x}{(x+1)} + c$ (D) $\frac{-e^x}{(x+1)^4} + c$

   ›Reveal solutionSolution

   The integrand fits the classic $\int e^x[f(x)+f'(x)]dx=e^xf(x)+c$ pattern with $f(x)=1/(x+1)^2$, giving the answer $\dfrac{e^x}{(x+1)^2}+c$ directly, with no further integration needed.

   Concept and Intuition

   Whenever an integrand has the shape $e^x$ times a sum of a function and its derivative, the antiderivative is simply $e^x$ times that function — because $\dfrac{d}{dx}\left[e^xf(x)\right]=e^x f(x)+e^xf'(x)=e^x[f(x)+f'(x)]$. Recognizing this pattern converts a seemingly hard rational-times-exponential integral into pure algebra: just guess $f(x)$ and check.

   Step-by-Step Solution

   1. Guess $f(x)=\dfrac{1}{(x+1)^2}$, motivated by the $(x+1)^3$ in the denominator (one power lower after differentiating).
   2. Compute $f'(x) = \dfrac{d}{dx}(x+1)^{-2} = -2(x+1)^{-3} = \dfrac{-2}{(x+1)^3}$.
   3. Form $f(x)+f'(x) = \dfrac{1}{(x+1)^2}-\dfrac{2}{(x+1)^3} = \dfrac{(x+1)-2}{(x+1)^3} = \dfrac{x-1}{(x+1)^3}$.
   4. This exactly matches the given integrand $\dfrac{x-1}{(x+1)^3}$. …

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7. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{1}{(x^4 + 1)^{5/4}}\, dx =$ (A) $-\dfrac{4}{(x^4 + 1)^{1/4}}$ (B) $\dfrac{1}{(x^4 + 1)^{1/4}}$ (C) $\dfrac{x}{(x^4 + 1)^{1/4}}$ (D) $-\dfrac{2}{(x^4 + 1)^{1/4}}$

   ›Reveal solutionSolution

   A power-pulling substitution ($t=1+x^{-4}$) reduces this to a simple power-rule integral; the answer is $\dfrac{x}{(x^4+1)^{1/4}}+c$.

   Concept and Intuition

   When an integrand is a power of $(x^n+1)$ with an awkward fractional exponent, factoring out $x^n$ from inside the bracket and substituting $t=1+x^{-n}$ often converts it into a clean power-rule integral in $t$.

   Step-by-Step Solution

   1. Write $x^4+1=x^4(1+x^{-4})$, so

   $(x^4+1)^{5/4}=x^5(1+x^{-4})^{5/4}.$

   2. The integrand becomes

   $\frac{1}{(x^4+1)^{5/4}}=x^{-5}(1+x^{-4})^{-5/4}.$

   3. Let $t=1+x^{-4}$, so $dt=-4x^{-5}\,dx\ \Rightarrow\ x^{-5}dx=-\dfrac{dt}{4}$.
   4. The integral becomes

   $\int -\frac14 t^{-5/4}\,dt=-\frac14\cdot\frac{t^{-1/4}}{-1/4}+c=t^{-1/4}+c.$

   5. Substitute back: $t^{-1/4}=(1+x^{-4})^{-1/4}=\left[\dfrac{x^4+1}{x^4}\right]^{-1/4}=\dfrac{x}{(x^4+1)^{1/4}}$. …

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8. AP EAPCET 2026Set eng-2026-05-18-FN1 markMCQ
   Q.If $\displaystyle\int \dfrac{x^2 - x + 1}{x^2 + 1}\, e^{\mathrm{Cot}^{-1}x}\, dx = A(x)\, e^{\mathrm{Cot}^{-1}x} + c$, then $A(x) =$ (A) $-x$ (B) $x$ (C) $\sqrt{1 - x}$ (D) $\sqrt{1 + x}$

   ›Reveal solutionSolution

   Match the integrand to the derivative pattern of $A(x)e^{\mathrm{Cot}^{-1}x}$; the function that works is $A(x)=x$.

   Concept and Intuition

   Integrals of the form $\int e^{\theta(x)}[\,\cdots\,]dx=A(x)e^{\theta(x)}+c$ are recognized by reverse-engineering the product rule: $\frac{d}{dx}[A(x)e^{\theta(x)}]=A'(x)e^{\theta(x)}+A(x)\theta'(x)e^{\theta(x)}$. Here $\theta(x)=\mathrm{Cot}^{-1}x$ has $\theta'(x)=-\dfrac1{1+x^2}$, so we need $A'(x)-\dfrac{A(x)}{1+x^2}$ to equal the given coefficient.

   Step-by-Step Solution

   1. Simplify the coefficient: $\dfrac{x^2-x+1}{x^2+1}=\dfrac{(x^2+1)-x}{x^2+1}=1-\dfrac{x}{x^2+1}$.
   2. We need $A(x)$ with

   $A'(x)-\frac{A(x)}{1+x^2}=1-\frac{x}{1+x^2}.$

   3. Try $A(x)=x$, so $A'(x)=1$:

   $1-\frac{x}{1+x^2}=1-\frac{x}{1+x^2}\ \checkmark$

   This matches exactly.

   4. So $\int\dfrac{x^2-x+1}{x^2+1}e^{\mathrm{Cot}^{-1}x}\,dx = x\,e^{\mathrm{Cot}^{-1}x}+c$, giving $A(x)=x$. …

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9. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
   Q.$\displaystyle\int \dfrac{e^{\sin x}(\sin 2x - 8\cos x)}{2(\sin x-3)^2}\,dx =$ (A) $e^{\sin x}(\sin x-3)+c$ (B) $\dfrac{e^{\sin x}}{(\sin x-3)^2}+c$ (C) $e^{\sin x}(\sin x-3)^2+c$ (D) $\dfrac{e^{\sin x}}{\sin x-3}+c$

   ›Reveal solutionSolution

   A "$\int e^{f(x)}[g(x)+g'(x)]$"-style integral in disguise — substitute $t=\sin x$ and recognize the resulting rational-times-exponential as the derivative of $\dfrac{e^t}{t-3}$. Answer: $\dfrac{e^{\sin x}}{\sin x-3}+c$.

   Concept and Intuition

   The identity $\displaystyle\int e^{u}\left[h(u)+h'(u)\right]du = e^u h(u)+c$ is the key trick behind many "exponential times trig" integrals. After substituting $t=\sin x$, the goal is to spot the integrand as $e^t\left[\dfrac1{t-3}-\dfrac1{(t-3)^2}\right]$, which is precisely $\dfrac{d}{dt}\!\left[\dfrac{e^t}{t-3}\right]$ by the quotient rule.

   Step-by-Step Solution

   1. Expand $\sin2x=2\sin x\cos x$ in the numerator:

   $e^{\sin x}(\sin2x-8\cos x) = e^{\sin x}\cos x(2\sin x-8) = 2e^{\sin x}\cos x(\sin x-4).$

   2. The integral becomes

   $\int \frac{2e^{\sin x}\cos x(\sin x-4)}{2(\sin x-3)^2}\,dx = \int \frac{e^{\sin x}\cos x(\sin x-4)}{(\sin x-3)^2}\,dx.$

   3. Substitute $t=\sin x \Rightarrow dt=\cos x\,dx$:

   $\int e^t\,\frac{t-4}{(t-3)^2}\,dt.$

   4. Decompose $\dfrac{t-4}{(t-3)^2}$ via partial fractions: write $t-4=A(t-3)+B$. At $t=3$: $B=-1$. Matching the $t$-coefficient: $A=1$. So $\dfrac{t-4}{(t-3)^2}=\dfrac{1}{t-3}-\dfrac{1}{(t-3)^2}$. …

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10. AP EAPCET 2025Set eng-2025-05-21-FN1 markMCQ
    Q.If $\displaystyle\int\left(3t^2\sin\dfrac{1}{t}-t\cos\dfrac{1}{t}\right)dt = f(t)\sin\left(\dfrac{1}{t}\right)+c$, then $f(2) =$ (A) $2$ (B) $-12$ (C) $8$ (D) $-16$

    ›Reveal solutionSolution

    Reverse-engineer $f(t)$ by matching the product-rule expansion of $f(t)\sin(1/t)$ to the given integrand term by term. Answer: $f(2)=8$.

    Concept and Intuition

    If $\displaystyle\int g(t)\,dt = f(t)\sin\!\left(\frac1t\right)+c$, then differentiating both sides must reproduce $g(t)$ exactly:

    $g(t) = f'(t)\sin\!\left(\frac1t\right) + f(t)\cos\!\left(\frac1t\right)\cdot\left(-\frac{1}{t^2}\right).$

    Matching the $\sin(1/t)$-coefficient and the $\cos(1/t)$-coefficient of the given $g(t)=3t^2\sin\frac1t - t\cos\frac1t$ to this template pins down $f$.

    Step-by-Step Solution

    1. Assume the antiderivative has the stated form $f(t)\sin(1/t)+c$; differentiate:

    $\frac{d}{dt}\left[f(t)\sin\frac1t\right] = f'(t)\sin\frac1t - \frac{f(t)}{t^2}\cos\frac1t.$

    2. This must equal $3t^2\sin\dfrac1t - t\cos\dfrac1t$.
    3. Match the $\sin(1/t)$ terms: $f'(t)=3t^2 \implies f(t)=t^3+C_0$.
    4. Match the $\cos(1/t)$ terms: $-\dfrac{f(t)}{t^2}=-t \implies f(t)=t^3$ — consistent with step 3 (so $C_0=0$).
    5. Thus $f(t)=t^3$, and $f(2)=2^3=8$. …

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11. AP EAPCET 2025Set eng-2025-05-21-AN1 markMCQ
    Q.If $\int e^{\sin x}(1 + \sec x \tan x)\,dx = e^{\sin x} f(x) + c$, then in $0 \le x \le 2\pi$, the number of solutions of $f(x) = 1$ is (A) $4$ (B) $0$ (C) $2$ (D) $3$

    ›Reveal solutionSolution

    Recognize the integrand as the derivative of $e^{\sin x}\sec x$, identify $f(x)=\sec x$, then count solutions of $\sec x=1$ on $[0,2\pi]$. Answer: $2$.

    Concept and Intuition

    Many integrals of the form $\int e^{g(x)}\big(g'(x)h(x) + h'(x)\big)dx$ are exact derivatives of $e^{g(x)}h(x)$ via the product rule. Spotting this pattern converts the integration problem into recognizing a product-rule expansion, rather than grinding through substitution.

    Step-by-Step Solution

    1. Try $h(x) = \sec x$ with $g(x) = \sin x$. Then

    $\frac{d}{dx}\left[e^{\sin x}\sec x\right] = e^{\sin x}\cos x \cdot \sec x + e^{\sin x}\cdot \sec x\tan x = e^{\sin x}\cdot 1 + e^{\sin x}\sec x\tan x = e^{\sin x}(1+\sec x\tan x).$

    2. This exactly matches the given integrand, so

    $\int e^{\sin x}(1+\sec x\tan x)\,dx = e^{\sin x}\sec x + c \implies f(x) = \sec x.$

    3. Solve $f(x)=1$: $\sec x = 1 \iff \cos x = 1 \iff x = 2n\pi$. …

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12. AP EAPCET 2025Set eng-2025-05-22-FN1 markMCQ
    Q.$\int\frac{\sin x + \cos x}{\sin x - \cos x}dx =$ (A) $-x + \log|\cos x - \sin x| + c$ (B) $x - \log|\cos x - \sin x| + c$ (C) $-\log|\cos x - \sin x| + c$ (D) $\log|\cos x - \sin x| + c$

    ›Reveal solutionSolution

    The numerator is exactly the derivative of the denominator, so this is a direct $\int f'(x)/f(x)\,dx=\log|f(x)|$ integral, giving $\log|\cos x-\sin x|+c$.

    Concept and Intuition

    Whenever the numerator of a rational trigonometric integrand is (up to a constant) the derivative of the denominator, the integral is immediately a logarithm — no partial fractions or trig identities needed.

    Step-by-Step Solution

    1. Let $u=\sin x-\cos x$. Then $\dfrac{du}{dx}=\cos x+\sin x$, which is exactly the numerator.
    2. So $\displaystyle\int\frac{\sin x+\cos x}{\sin x-\cos x}\,dx=\int\frac{du}{u}=\log|u|+c=\log|\sin x-\cos x|+c$.
    3. Since $|\sin x-\cos x|=|-(\cos x-\sin x)|=|\cos x-\sin x|$, this is the same as $\log|\cos x-\sin x|+c$. …

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